Differentiation finds the gradient of a curve at any point. To differentiate axⁿ, multiply by the power and reduce the power by one, giving anxⁿ⁻¹; a constant differentiates to 0. Substitute an x-value into dy/dx to find the gradient there. At a turning point the gradient is zero, so solve dy/dx = 0 to find it.
Differentiation rules
- y = axⁿ gives dy/dx = anxⁿ⁻¹
- y = ax gives dy/dx = a; a constant gives 0
- Differentiate each term separately
- Gradient at a point: substitute x into dy/dx
- Turning points: solve dy/dx = 0
What differentiation tells you
The derivative dy/dx is a formula for the gradient of a curve. Unlike a straight line, a curve's gradient changes from point to point, and dy/dx gives the gradient at any x-value.
Differentiating powers of x
Multiply by the power, then reduce the power by one. So x³ becomes 3x², 5x² becomes 10x, 4x becomes 4, and a constant becomes 0. Differentiate each term of a polynomial separately: y = x³ + 2x² − 5x + 1 gives dy/dx = 3x² + 4x − 5.
Gradient at a point
Differentiate first, then substitute the x-coordinate into dy/dx. For y = 2x³ − x, dy/dx = 6x² − 1, so at x = 2 the gradient is 6 × 4 − 1 = 23. Substituting into y instead would give the height of the curve, not its gradient.
Turning points
At a turning point the curve is momentarily flat, so dy/dx = 0. Solve that equation for x, then substitute back into the original equation for y. For y = x² + 8x + 11, dy/dx = 2x + 8 = 0 gives x = −4 and y = −5, so the turning point is (−4, −5).
Maximum or minimum?
A U-shaped turning point is a minimum and an upside-down U is a maximum. One way to decide is to check the gradient just either side: negative then positive means a minimum, positive then negative means a maximum. For a cubic, there can be two turning points, one of each type.
Worked examples
Example 1. Differentiate y = x³ + 2x² − 5x + 1.
- x³ → 3x²; 2x² → 4x; −5x → −5; 1 → 0.
Answer: dy/dx = 3x² + 4x − 5.
Example 2. Find the gradient of y = 2x³ − x at x = 2.
- dy/dx = 6x² − 1.
- At x = 2: 6 × 4 − 1.
Answer: 23.
Example 3. Find the turning point of y = x² + 8x + 11.
- dy/dx = 2x + 8 = 0, so x = −4.
- y = 16 − 32 + 11 = −5.
Answer: (−4, −5).
Example 4. Find the turning points of y = 48x − x³ and say which is a maximum.
- dy/dx = 48 − 3x² = 0, so x² = 16 and x = ±4.
- x = 4: y = 192 − 64 = 128. x = −4: y = −192 + 64 = −128.
- The gradient changes from positive to negative at x = 4.
Answer: Maximum (4, 128); minimum (−4, −128).
Example 5. A ball's height is h = 20t − 5t² metres after t seconds. Find its greatest height.
- dh/dt = 20 − 10t = 0, so t = 2.
- h = 40 − 20 = 20.
Answer: 20 m, after 2 seconds.
Common mistakes
- Forgetting to reduce the power. x³ differentiates to 3x², not 3x³.
- Not differentiating the constant to 0. A constant term disappears.
- Substituting into y for a gradient. The gradient comes from dy/dx.
- Stopping at the x-coordinate. A turning point needs both coordinates.
- Differentiating 4x to 4x. 4x differentiates to 4.
How differentiation is examined
Differentiation is part of the Higher tier of the Edexcel International GCSE (4MA1). It is not in the UK Edexcel GCSE (1MA1), where gradients of curves are estimated with tangents instead. iGCSE questions ask you to:
- differentiate a polynomial
- find the gradient of a curve at a given point
- find the coordinates of a turning point
- solve a maximum or minimum problem, such as greatest height or area
The GCSE algebra revision engine has differentiation practice, and gradients and rates of change covers the graphical side.
Differentiation continues at A-Level in basic differentiation and stationary points.
Frequently asked questions
How do you differentiate xⁿ?
Multiply by n and reduce the power by 1: the derivative is nxⁿ⁻¹.
What does dy/dx = 0 mean?
The curve is flat there, which happens at a turning point.
Is differentiation on the GCSE syllabus?
It is in the Edexcel International GCSE (4MA1) Higher tier, but not in the UK GCSE (1MA1).